The basic notions from linear algebra are assumed to be well-known and so they are not recalled here. However, we briefly give again the definition of vector space and fix some general terminology for linear mappings between vector spaces. In this section we are going to consider vector spaces over the fieldKof real or complex numbers which is given the usual euclidean topology defined by means of the modulus.
Definition 1.2.1. A setX with the two mappings:
X×X → X
(x, y) 7→ x+y vector addition K×X → X
(λ, x) 7→ λx scalar multiplication
is a vector space(or linear space) overKif the following axioms are satisfied:
(L1) 1. (x+y) +z=x+ (y+z),∀x, y, z∈X (associativity of +) 2. x+y=y+x,∀x, y∈X (commutativity of +)
3. ∃o∈X: x+o=x,∀x,∈X (neutral element for +)
4. ∀x∈X, ∃! −x∈X s.t. x+ (−x) =o (inverse element for +) (L2) 1. λ(µx) = (λµ)x, ∀x∈X, ∀λ, µ∈K
(compatibility of scalar multiplication with field multiplication) 2. 1x=x∀x∈X (neutral element for scalar multiplication) 3. (λ+µ)x=λx+µx, ∀x∈X, ∀λ, µ∈K
(distributivity of scalar multiplication with respect to field addition) 4. λ(x+y) =λx+λy,∀x, y∈X, ∀λ∈K
(distributivity of scalar multiplication wrt vector addition) Definition 1.2.2.
Let X, Y be two vector space over K. A mapping f : X → Y is called lin-ear mapping or homomorphism if f preserves the vector space structure, i.e.
f(λx+µy) =λf(x) +µf(y)∀x, y∈X, ∀λ, µ∈K.
Definition 1.2.3.
• A linear mapping from X to itself is called endomorphism.
• A one-to-one linear mapping is called monomorphism. IfS is a subspace of X, the identity map is a monomorphism and it is called embedding.
• An onto (surjective) linear mapping is called epimorphism.
• A bijective (one-to-one and onto) linear mapping between two vector spaces X and Y over K is called (algebraic) isomorphism. If such a map exists, we say that X andY are (algebraically) isomorphicX∼=Y.
• An isomorphism from X into itself is called automorphism.
It is easy to prove that: A linear mapping is one-to-one (injective) if and only if f(x) = 0 implies x= 0.
Definition 1.2.4. A linear mapping from X→ K is called linear functional or linear formonX. The set of all linear functionals onX is called algebraic dual and it is denoted by X∗.
Note that the dual space of a finite dimensional vector spaceX is isomor-phic to X.
Topological Vector Spaces
2.1 Definition and properties of a topological vector space
In this section we are going to consider vector spaces over the field Kof real or complex numbers which is given the usual euclidean topology defined by means of the modulus.
Definition 2.1.1. A vector spaceX overKis called a topological vector space (t.v.s.) if X is provided with a topology τ which is compatible with the vector space structure of X, i.e.τ makes the vector space operations both continuous.
More precisely, the condition in the definition of t.v.s. requires that:
X×X → X
(x, y) 7→ x+y vector addition K×X → X
(λ, x) 7→ λx scalar multiplication
are both continuous when we endow X with the topology τ,K with the eu-clidean topology,X×XandK×Xwith the correspondent product topologies.
Remark 2.1.2. If (X, τ) is a t.v.s then it is clear from Definition2.1.1 that PN
k=1λ(n)k x(n)k → PN
k=1λkxk as n → ∞ w.r.t. τ if for each k = 1, . . . , N as n → ∞ we have that λ(n)k → λk w.r.t. the euclidean topology on K and x(n)k →xk w.r.t. τ.
Let us discuss now some examples and counterexamples of t.v.s.
Examples 2.1.3.
a) Every vector space X over K endowed with the trivial topology is a t.v.s..
b) Every normed vector space endowed with the topology given by the metric induced by the norm is a t.v.s. (see Exercise Sheet 1).
c) There are also examples of spaces whose topology cannot be induced by a norm or a metric but that are t.v.s., e.g. the space of infinitely differen-tiable functions, the spaces of test functions and the spaces of distributions endowed with suitable topologies (which we will discuss in details later on).
In general, a metric vector space is not a t.v.s.. Indeed, there exist metrics for which both the vector space operations of sum and product by scalars are discontinuous (see Exercise Sheet 1 for an example).
Proposition 2.1.4. Every vector space X over K endowed with the discrete topology is not a t.v.s. unless X={o}.
Proof. Assume that it is a t.v.s. and take o6=x ∈ X. The sequence αn = n1 in K converges to 0 in the euclidean topology. Therefore, since the scalar multiplication is continuous, αnx → o by Proposition 1.1.39, i.e. for any neighbourhood U ofoinX there existsm∈Ns.t. αnx∈U for alln≥m. In particular, we can takeU ={o} since it is itself open in the discrete topology.
Hence,αmx=o, which implies that x=oand so a contradiction.
Definition 2.1.5. Two t.v.s. XandY overKare (topologically) isomorphic if there exists a vector space isomorphism X → Y which is at the same time a homeomorphism (i.e. bijective, linear, continuous and inverse continuous).
In analogy to Definition 1.2.3, let us collect here the corresponding termi-nology for mappings between two t.v.s..
Definition 2.1.6. Let X and Y be two t.v.s. on K.
• A topological homomorphism f from X to Y is a continuous linear mapping which is also open, i.e. every open set in X is mapped to an open set in f(X) (endowed with the subset topology induced by Y).
• A topological monomorphism from X to Y is an injective topological homomorphism.
• A topological isomorphismfrom X to Y is a bijective topological homo-morphism.
• A topological automorphism of X is a topological isomorphism from X into itself.
Proposition 2.1.7. Given a t.v.s. X, we have that:
1. For any x0 ∈ X, the mapping x 7→ x +x0 (translation by x0) is a homeomorphism of X onto itself.
2. For any 06=λ∈K, the mapping x7→λx(dilationby λ) is a topological automorphism ofX.
Proof. Both mappings are continuous as X is a t.v.s.. Moreover, they are bijections by the vector space axioms and their inverses x 7→ x−x0 and x 7→ λ1x are also continuous. Note that the second map is also linear so it is a topological automorphism.
Proposition2.1.7–1 shows that the topology of a t.v.s. is always a transla-tion invariant topology, i.e. all translatransla-tions are homeomorphisms. Note that the translation invariance of a topologyτ on a vector spaceXis not sufficient to conclude (X, τ) is a t.v.s..
Example 2.1.8. If a metric d on a vector space X is translation invariant, i.e. d(x+z, y+z) = d(x, y) for all x, y, z ∈ X (e.g. the metric induced by a norm), then the topology induced by the metric is translation invariant and the addition is always continuous. However, the multiplication by scalars does not need to be necessarily continuous (take d to be the discrete metric, then the topology generated by the metric is the discrete topology which is not compatible with the scalar multiplication see Proposition 2.1.4).
The translation invariance of the topology of a t.v.s. means, roughly speak-ing, that a t.v.s. X topologically looks about any point as it does about any other point. More precisely:
Corollary 2.1.9. The filter F(x) of neighbourhoods of x in a t.v.s. X coin-cides with the family of the setsO+xfor allO∈ F(o), whereF(o)is the filter of neighbourhoods of the origin o (i.e. neutral element of the vector addition).
Proof. (Exercise Sheet 1)
Thus the topology of a t.v.s. is completely determined by the filter of neighbourhoods of any of its points, in particular by the filter of neighbour-hoods of the origin oor, more frequently, by a base of neighbourhoods of the origin o. Therefore, we need some criteria on a filter of a vector space X which ensures that it is the filter of neighbourhoods of the origin w.r.t. some topology compatible with the vector structure of X.
Theorem 2.1.10. A filter F of a vector space X over K is the filter of neighbourhoods of the origin w.r.t. some topology compatible with the vector structure of X if and only if
1. The origin belongs to every setU ∈ F 2. ∀U ∈ F, ∃V ∈ F s.t. V +V ⊂U
3. ∀U ∈ F, ∀λ∈K withλ6= 0 we have λU ∈ F 4. ∀U ∈ F, U is absorbing.
5. ∀U ∈ F, ∃V ∈ F balanced s.t. V ⊂U.
Before proving the theorem, let us fix some definitions and notations:
Definition 2.1.11. Let U be a subset of a vector space X.
1. U is absorbing(or radial) if∀x∈X ∃ρ >0 s.t. ∀λ∈Kwith|λ| ≤ρwe have λx∈ U. Roughly speaking, we may say that a subset is absorbing if it can be made by dilation to swallow every point of the whole space.
2. U is balanced (or circled) if ∀x ∈ U, ∀λ ∈ K with |λ| ≤ 1 we have λx ∈ U. Note that the line segment joining any point x of a balanced setU to −x lies in U.
Clearly,omust belong to every absorbing or balanced set. The underlying field can make a substantial difference. For example, if we consider the closed interval [−1,1]⊂R then this is a balanced subset of C as real vector space, but if we take C as complex vector space then it is not balanced. Indeed, if we take i∈C we get thati1 =i /∈[−1,1].
Examples 2.1.12.
a) In a normed space the unit balls centered at the origin are absorbing and balanced.
b) The unit ball B centered at(12,0)∈R2 is absorbing but not balanced in the real vector space R2. Indeed, B is a neighbourhood of the origin and so by Theorem2.1.10-4 is absorbing. However,B is not balanced because for example if we take x= (1,0)∈B and λ=−1 then λx /∈B.
c) In the real vector space R2 endowed with the euclidean topology, the subset in Figure 2.1 is absorbing and the one in Figure 2.2is balanced.
Figure 2.1: Absorbing Figure 2.2: Balanced
d) The polynomials R[x] are a balanced but not absorbing subset of the real space C([0,1],R) of continuous real valued functions on [0,1]. Indeed, any multiple of a polynomial is still a polynomial but not every continuous function can be written as multiple of a polynomial.
e) The subset A := {(z1, z2) ∈ C2 :|z1| ≤ |z2|} of the complex space C2 en-dowed with the euclidean topology is balanced butA˚is not balanced. Indeed,
∀(z1, z2)∈A and ∀λ∈Cwith |λ| ≤1 we have that
|λz1|=|λ||z1| ≤ |λ||z2|=|λz2|
i.e. λ(z1, z2) ∈ A. Hence, A is balanced. If we consider instead ˚A = {(z1, z2) ∈ C2 : |z1| < |z2|} then ∀(z1, z2) ∈ A˚and λ = 0 we have that λ(z1, z2) = (0,0)∈/ A. Hence, ˚˚ A is not balanced.
Proposition 2.1.13.
a) If B is a balanced subset of a t.v.s. X then so isB¯.
b) If B is a balanced subset of a t.v.s. X and o∈B˚ thenB˚is balanced.
Proof. (Exercise Sheet 1)
Combining this result and Theorem2.1.10, we can easily obtain that:
Corollary 2.1.14.
a) Every t.v.s. has always a base of closed neighbourhoods of the origin.
b) Every t.v.s. has always a base of balanced absorbing neighbourhoods of the origin. In particular, it has always a base of closed balanced absorbing neighbourhoods of the origin.
c) Proper subspaces of a t.v.s. are never absorbing. In particular, if M is an open subspace of a t.v.s. X then M =X.
Proof. (Exercise Sheet 1) Proof. of Theorem2.1.10.
Necessity part.
Suppose that X is a t.v.s. then we aim to show that the filter of neighbour-hoods of the origin F satisfies the properties 1,2,3,4,5. LetU ∈ F.
1. obvious, since every set U ∈ F is a neighbourhood of the origino.
2. Since by the definition of t.v.s. the addition (x, y)7→x+yis a continuous mapping, the preimage ofU under this map must be a neighbourhood of (o, o)∈X×X. Therefore, it must contain a rectangular neighbourhood W ×W0 whereW, W0 ∈ F. TakingV =W ∩W0 we get the conclusion, i.e. V +V ⊂U.
3. By Proposition 2.1.7, fixed an arbitrary 06=λ∈K, the map x7→ λ−1x ofXinto itself is continuous. Therefore, the preimage of any neighbour-hoodU of the origin must be also such a neighbourhood. This preimage is clearly λU, henceλU ∈ F.
4. Let x ∈X. By the continuity of the scalar multiplication at the point (0, x), the preimage of U under this map must contain a rectangular neighbourhood N ×(x+W) where N is a neighbourhood of 0 in the euclidean topology onKandW ∈ F. Hence, there existsρ >0 such that Bρ(0) :={α ∈ K: |α| ≤ ρ} ⊆ N. ThusBρ(0)×(x+W) is contained in the preimage ofU under the scalar multiplication, i.e. for anyy∈W and any λ∈ K with |λ| ≤ ρ we have λ(x+y) ∈ U. In particular, for y=o we get that U is absorbing.1
5. By the continuity of the scalar multiplication at the point (0, o)∈K×X, the preimage ofU under this map must contain a rectangular neighbour-hoodN×W whereN is a neighbourhood of 0 in the euclidean topology on K and W ∈ F. On the other hand, there exists ρ > 0 such that Bρ(0) := {α ∈ K : |α| ≤ ρ} ⊆ N. Thus Bρ(0)×W is contained in the preimage of U under the scalar multiplication, i.e. αW ⊂U for all α ∈K with|α| ≤ρ. Hence, the set V =S
|α|≤ραW ⊂U. Now V ∈ F since each αW ∈ F by 3 and V is clearly balanced (since for anyx∈V there existsα∈Kwith|α| ≤ρs.t. x∈αW and therefore for anyλ∈K with|λ| ≤1 we getλx∈λαW ⊂V because|λα| ≤ρ).
Sufficiency part.
Suppose that the conditions 1,2,3,4,5 hold for a filterF of the vector spaceX.
We want to show that there exists a topologyτ onX such that F is the filter of neighbourhoods of the origin w.r.t. to τ and (X, τ) is a t.v.s. according to Definition2.1.1.
Let us define for any x∈X the filterFx :={U+x: U ∈ F }. It is easy to see that Fx fulfills the properties (N1) and (N2) of Theorem 1.1.10. In fact, we have:
• By 1 we have that∀U ∈ F, o∈U, then∀U ∈ F, x=o+x∈U+x, i.e.
∀A∈ Fx, x∈A.
• LetA∈ Fx then A=U +x for someU ∈ F. By 2, we have that there exists V ∈ F s.t. V +V ⊂U. Define B := V +x ∈ Fx and take any y ∈ B then we have V +y ⊂ V +B ⊂V +V +x ⊂U +x =A. But V +y belongs to the filterFy and therefore so doesA.
By Theorem1.1.10, there exists a unique topologyτ onX such thatFx is the filter of neighbourhoods ofx∈Xand so for which in particularF is the filter
1Alternative proof Suppose thatU is not absorbing. Then there existsx∈X such that for anyn∈Nthere existsλn∈K with|λn| ≤ n1 andλnx /∈U. But the sequenceλn→0 as n → ∞ and so the continuity of the scalar multiplication provides that also λnx→ o as n → ∞. Therefore, U contains infinitely many elements of {λnx}n∈N which yields a contradiction.
of neighbourhoods of the origin (i.e. Fx ≡ Fτ(x),∀x ∈ X and in particular F ≡ Fτ(o)).
It remains to prove that the vector addition and the scalar multiplication inX are continuous w.r.t. toτ.
• The continuity of the addition easily follows from the property 2. Indeed, let (x0, y0)∈X×X and take a neighbourhood W of its image x0+y0. Then W =U +x0+y0 for some U ∈ F. By 2, there existsV ∈ F s.t.
V +V ⊂ U and so (V +x0) + (V +y0) ⊂ W. This implies that the preimage of W under the addition contains (V +x0)×(V +y0) which is a neighbourhood of (x0, y0).
• To prove the continuity of the scalar multiplication, let (λ0, x0)∈K×X and take a neighbourhood U0 of λ0x0. Then U0 = U +λ0x0 for some U ∈ F. By 2 and 5, there exists W ∈ F s.t. W +W +W ⊂U and W is balanced. By 4, W is also absorbing so there existsρ >0 (w.l.o.g. we can takeρ≤1 because of property 3) such that ∀λ∈Kwith|λ| ≤ρwe have λx0 ∈W.
Supposeλ0 = 0 thenλ0x0 =o andU0 =U. Now
Im(Bρ(0)×(W +x0)) ={λy+λx0 : λ∈Bρ(0), y∈W}.
As λ ∈ Bρ(0) and W is absorbing, λx0 ∈ W. Also since |λ| ≤ ρ ≤ 1 for all λ ∈ Bρ(0) and since W is balanced, we have λW ⊂ W. Thus Im(Bρ(0)×(W+x0))⊂W+W ⊂W+W+W ⊂U and so the preimage ofU under the scalar multiplication containsBρ(0)×(W+x0) which is a neighbourhood of (0, x0).
Supposeλ06= 0 and takeσ= min{ρ,|λ0|}. ThenIm((Bσ(0) +λ0)× (|λ0|−1W +x0))={λ|λ0|−1y+λx0+λ0|λ0|−1y+λx0 :λ∈Bσ(0), y∈W}.
As λ ∈ Bσ(0), σ ≤ ρ and W is absorbing, λx0 ∈ W. Also since ∀λ ∈ Bσ(0) the modulus of λ|λ0|−1 and λ0|λ0|−1 are both ≤1 and since W is balanced, we have λ|λ0|−1W, λ0|λ0|−1W ⊂ W. Thus Im(Bσ(0) + λ0×(|λ0|−1W +x0)) ⊂ W +W +W +λ0x0 ⊂ U +λ0x0 and so the preimage of U +λ0x0 under the scalar multiplication contains Bσ(0) + λ0×(|λ0|−1W +x0) which is a neighbourhood of (λ0, x0).
Let us show some further useful properties of the t.v.s.:
Proposition 2.1.15.
1. Every linear subspace of a t.v.s. endowed with the correspondent subspace topology is itself a t.v.s..
2. The closure H of a linear subspace H of a t.v.s. X is again a linear subspace of X.
3. Let X, Y be two t.v.s. and f :X→ Y a linear map. f is continuous if and only if f is continuous at the origin o.
Proof.
1. This clearly follows by the fact that the addition and the multiplication restricted to the subspace are just a composition of continuous maps (recall that inclusion is continuous in the subspace topology c.f. Defini-tion 1.1.19).
2. Let x0, y0 ∈ H and take any U ∈ F(o). By Theorem 2.1.10-2, there existsV ∈ F(o) s.t. V +V ⊂U. Then, by definition of closure points, there existx, y∈H s.t. x∈V+x0andy∈V+y0. Therefore,x+y∈H (sinceHis a linear subspace) andx+y∈(V+x0)+(V+y0)⊂U+x0+y0. Hence, x0+y0 ∈H. Similarly, one can prove that if x0 ∈H,λx0 ∈ H for any λ∈K.
3. Assume thatf is continuous ato∈X and fix anyx6=oinX. Let U be an arbitrary neighbourhood of f(x)∈ Y. By Corollary2.1.9, we know that U = f(x) +V where V is a neighbourhood of o ∈ Y. Since f is linear we have that: f−1(U) = f−1(f(x) +V) ⊃ x+f−1(V). By the continuity at the origin of X, we know thatf−1(V) is a neighbourhood of o∈X and so x+f−1(V) is a neighbourhood of x∈X.